If the line (x + g) cos theta + (y + f) sin t | vedclass

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(C) The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.The given line is $(x + g) \co

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$g^2 + f^2 = k^2 + c$

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(C) The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.The given line is $(x + g) \cos 	heta + (y + f) \sin 	heta = k$,which can be rewritten as $x \cos 	heta + y \sin 	heta + (g \cos 	heta + f \sin 	heta - k) = 0$.If the line is tangent to the circle,the perpendicular distance from the center $C(-g, -f)$ to the line must equal the radius $r$.$r = \left| \frac{(-g) \cos 	heta + (-f) \sin 	heta + g \cos 	heta + f \sin 	heta - k}{\sqrt{\cos^2 	heta + \sin^2 	heta}} ight|$.Simplifying the numerator,we get $r = \left| \frac{-k}{1} ight| = |k|$.Thus,$\sqrt{g^2 + f^2 - c} = |k|$.Squaring both sides,we get $g^2 + f^2 - c = k^2$,which simplifies to $g^2 + f^2 = k^2 + c$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

If the line $(x + g) \cos \theta + (y + f) \sin \theta = k$ is tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,then:

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